I was trying to demonstrate geometricaly that the round shaded parts on the bottom and the left were the missing part of the shaded area respectively on the top and on the right. I think I lacked a symetry argument (or something to apply Thales) to show the circle cuts the square at the same place on each side. You ended up demonstrating what I wanted but with subtracting shapes. Clever, I think.

Hi, could you possibly briefly explain how you know the line at 1:55 is a diameter? Thanks!

@ThePhantomoftheMath

27 күн бұрын

Sure: Our purple triangle is a right triangle, meaning it has a 90-degree angle. Since we have inscribed a 90-degree angle, the arc that it subtends must be a 180-degree arc. Therefore, the hypotenuse of the purple triangle must be the diameter of the circle.

how do u have only 287 subs wtffffff

@ThePhantomoftheMath

26 күн бұрын

Lol. Thank you for that! :) I only started a couple of months ago, so I'm relatively new to this. But your comment gives me the energy to keep making more and more videos like this.

Excellent way of manipulation of areas

@ThePhantomoftheMath

11 күн бұрын

Thank you!

Hi, what an interesting lesson !!! Havn‘t yet heard of the surprising Formular 4r^2 😳 A mathematical treasure 👍🏽😊👍🏽

@ThePhantomoftheMath

17 күн бұрын

Hi! I'm glad you liked the video. I was also surprised by this formula when I found out about it, so I knew I needed to make a video that includes it. It's truly amazing.

At 2:05 how do you know the purple triangle is isoceles?

@ThePhantomoftheMath

20 күн бұрын

The purple triangle is isosceles for the following reasons: - The square is inscribed in the circle, meaning the circle passes through all four vertices of the square. Therefore, the diameter of the circle is equal to the diagonal of the square. - The green triangle's vertices are: The top-left corner of the square. The bottom-right corner of the square. The bottom-left corner of the square. - The line from the bottom-left vertex to the top-left vertex of the square is a side of the square, which we can call - a. - The diagonal of the square (which is the hypotenuse of the right triangle formed by the square's sides) has a length of - a root of 2. - In the purple triangle, the two sides from the bottom-left vertex to the top-left vertex and from the bottom-left vertex to the bottom-right vertex are equal in length because they are both sides of the square, each with length - a. Therefore, the purple triangle is isosceles because it has two sides of equal length.

Nice!

Points to be labeled: Center of the circle: O Vertices of the square (clockwise from top left): ABCD. Circle/square intersection/tangent points: M (tangent to AB) N (tangent to BC) P (intersects CD) Q (intersects DA) Let r be the radius of the circle. Draw diagonal DB. Let the point on the circumference that it passes through be T. As the figure is symmetrical about DB, it will pass through O, thus OD = OT = r. Draw QP. As ponts Q, P, and D are all points on the circumference of a circle, and ∠PDQ = 90° (as the vertex of a square), then QP must be a diameter of the circle, by Thales' Theorem. Draw chords QT and TP. As DT is already known to be a diameter of the circle, and QT and TP are chords and thus perpendicular to any radii passing through their midpoints, they are parallel to tangents AB and BC respectively and DQTP is a square. As diagonals DT and QP are diameters of circle O, their lengths are 2r. Thus the side length s of DQTP is 2r/√2 or √2r. This means that the thicknesses of the circular segments subtended by arcPD and ard DQ, and by symmetry the distance from QT to AB and TP to BC, is (2r-√2r)/2 = (2-√2)r/2. Thus the side length of the larger square is √2r plus this amount. 1 = √2r + (2-√2)r/2 1 = √2r + r - r/√2 1 = (2+√2-1)r/√2 r = √2/(√2+1) r = √2(√2-1)/(√2+1)(√2-1) r = (2-√2)/(2-1) r = 2 - √2 s = √2r = √2(2-√2) = 2√2 - 2 The area of the shaded region is equal ro the difference between the areas of the two squares, as the chords QT and TP are the same distance from the center of the circle as the chords PD and DQ that form the two orange circular segments. Area = 1² - (2√2-2)² A = 1 - (8-8√2+4) A = 8√2 - 11 ≈ 0.314

Con centro el del círculo, giramos 180º los dos segmentos circulares; en la nueva posición, sus cuerdas delimitan un nuevo cuadrado inscrito en el círculo, cuyo vértice inferior izquierdo y los lados y diagonal correspondiente se superponen a los del cuadrado de lado 1 ud → En el trazado resultante, los dos segmentos que unen el centro del círculo con los puntos de tangencia y el que lo une con el vértice común de ambos cuadrados tienen una longitud igual al radio "r" del círculo.→ r+r√2=1*√2→ r(1+√2)=√2→ r=2-√2 → Área roja =Diferencia entre las áreas de ambos cuadrados → 1²-[(2r)²/2] =1-2r² =1-2(2-√2)² =8√2 -11. Gracias y un saludo cordial.

was expecting pi, left dissapoinyed. but great work!

@ThePhantomoftheMath

27 күн бұрын

Sorry for that! Next time will be Pi included 😂

@Larsbutb4d

27 күн бұрын

@@ThePhantomoftheMath its ok js expected pi bc yknow... circle